Handbook of Civil Engineering Calculations

STRESS CAUSED BY AN AXIAL LOAD
A concentric load of 20,000 lb (88,960 N) is applied to a hanger having a cross-sectional
area of 1.6 sq.in. (1032.3 mm 2 ). What is the axial stress in the hanger?
Calculation Procedure:
1. Compute the axial stress
Use the general stress relation s P/A 20,000/1.6 12,500 lb/sq.in. (86,187.5 kPa).
Related Calculations. Use this general stress relation for a member of any crosssectional
shape, provided the area of the member can be computed and the member is
made of only one material.
DEFORMATION CAUSED BY AN
AXIAL LOAD
A member having a length of 16 ft (4.9 m) and a cross-sectional area of 2.4 sq.in. (1548.4
mm 2 ) is subjected to a tensile force of 30,000 lb (133.4 kN). If E 15 10 6 lb/sq.in.
(103 GPa), how much does this member elongate?
Calculation Procedure:
1. Apply the general deformation equation
The general deformation equation is l PL/(AE) 30,000(16)(12)/[2.4(15 10 6 )1
0.16 in. (4.06 mm).
Related Calculations. Use this general deformation equation for any material
whose modulus of elasticity is known. For composite materials, this equation must be altered
before it can be used.
DEFORMATION OF A BUILT-UP MEMBER
A member is built up of three bars placed end to end, the bars having the lengths and crosssectional
areas shown in Fig. 16. The member is placed between two rigid surfaces and axial
loads of 30 kips (133 kN) and 10 kips (44 kN) are applied at A and B, respectively. If E
2000 kips/sq.in. (13,788 MPa), determine the horizontal displacement of A and B.
Calculation Procedure:
STATICS, STRESS AND STRAIN, AND FLEXURAL ANALYSIS
1. Express the axial force in terms of one reaction
Let R L and R R denote the reactions at the left and right ends, respectively. Assume that
both reactions are directed to the left. Consider a tensile force as positive and a compressive
force as negative. Consider a deformation positive if the body elongates and negative
if the body contracts.
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